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Q.

On a given condition, the equilibrium concentration of HI, H2 and I2 are 0.80, 0.10 and 0.10 mole/litre. The equilibrium constant for the reaction H2+ I2  ⇋ 2HI will be

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a

64

b

12

c

8

d

0.8

answer is A.

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Detailed Solution

H2 + I2 ⇋ 2HI;  HI = 0.80,  H2 = 0.10,  I2 = 0.10 KC=HI2H2 I2= 0.80 × 0.800.10 × 0.10=64
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